Azzera filtri
Azzera filtri

Find and replace values in a cell array containing 3-D matrixes with numeric values

1 visualizzazione (ultimi 30 giorni)
Hi, I have a cell array with 6 cells, each cell containing a 3-D matrix of 640x480x30. I'm trying to find all zeros (0) in the cell array and replace them with NaNs.
In the general case, I have a cell array (MyCellArray) with -K- cells, each cell containing an arbitrary vector or matrix in an arbitrary size, with numeric values. I'm trying to find all places where MyCellArray contains the value of X (some number) and replace it with Y (some number).
Is there a way to do it without a for loop? Something like (that of course can't work) MyCellArray{MyCellArray==0}=NaN
Thanks!

Risposta accettata

Akira Agata
Akira Agata il 20 Apr 2018

Another possible solution:

YourCellArray = cellfun(@replaceZeroWithNan,YourCellArray,'UniformOutput',false);
function D = replaceZeroWithNan(D)
idx = D == 0;
D(idx) = nan;
end
  3 Commenti
Avishay Assayag
Avishay Assayag il 20 Apr 2018
Thanks for your answer! That's an excellent solution for the general case. Kudos!
Guillaume
Guillaume il 20 Apr 2018
Modificato: Guillaume il 20 Apr 2018

Note that, while more concise, cellfun is generally slower than an explicit loop.

Accedi per commentare.

Più risposte (1)

Guillaume
Guillaume il 20 Apr 2018
Modificato: Guillaume il 20 Apr 2018

In the generic case, it is not possible to do it without a loop (or cellfun, but in this case, you'd have to use a .m function not an anonymous function:

for idx = 1:numel(yourcellarray)
   m = yourcellarray{idx};
   m(m == 0) = NaN;
   yourcellarray{idx} = m;
end

In your example case, where all the matrices are the same size, then you'd be better off not using a cell array but a 4-D matrix. The replacement is then trivial:

m = cat(4, yourcellarray{:});
m(m == 0) = NaN;
  5 Commenti

Accedi per commentare.

Categorie

Scopri di più su Cell Arrays in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by